My question relates to MTW's Gravitation, Box 11.2 (copied below) and the discussion on page 271. This is my paraphrase of the gist of box 11.2:

Consider a family $\Lambda$ of timelike geodesics affinely parameterized by $\lambda$ and selected by the parameter $n$. The fiducial geodesic is designated by the selector value $n$. The set of ordered pairs $\left\langle \lambda+\Delta\lambda,n+\Delta n\right\rangle $ may serve as coordinates near $\mathscr{M}.$ In these coordinates, $\mathscr{M}$ has coordinates $\left\langle \lambda,n\right\rangle $. Define the vectors $\mathbf{u}=\frac{\partial}{\partial\lambda}$ and $\mathbf{n}=\frac{\partial}{\partial n}$. The separation between the fiducial point $\left\langle \lambda,n\right\rangle $ and the point with the same $\lambda$ value on the geodesic designated by $n+\Delta n$ is $\vec{n}=\Delta n\mathbf{n}$. The separation $\vec{n}$ is then parallel transported along the geodesic $n$ by $\Delta\lambda,$ so that the tail of its image is at $\mathscr{N}$ and the tip is at $\mathscr{B}$. A similar image is produced by parallel transport of $\vec{n}$ along the geodesic $n$ by $-\Delta\lambda,$ with tail and tip designated $\mathscr{L}$ and $\mathscr{A}$ respectively.

Beginning at the point $\mathscr{Q}$ which has coordinates $\left\langle \lambda,n+\Delta n\right\rangle$ the point $\mathscr{R}$ is determined by a parameter change $\Delta\lambda$ along the geodesic $n+\Delta n.$ The point $\mathscr{P}$ is determined by a change $-\Delta\lambda$ along the same $n+\Delta n$ geodesic.

The points $\mathscr{A}$ and $\mathscr{B}$ are determined by parallel transporting $\vec{n}=\Delta n\mathbf{n}$ along the geodesic $n$ by $-\Delta\lambda$ and $\Delta\lambda$ respectively. The vectors $\mathscr{B}\mathscr{R}$ and $\mathscr{A}\mathscr{P}$ are then parallel transported (along unspecified routs) to $\mathscr{Q}$ were they are summed to produce $$ \delta_{2}=\mathscr{B}\mathscr{R}+\mathscr{A}\mathscr{P}=\left(\Delta\lambda\right)^{2}\Delta n\left(\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n}\right), $$

where $\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n}$ is defined to be the relative-acceleration vector.

On page 271 we find:

[Examine Box 11.4] Thereby arrive at the remarkable equation (11.6)

$$ \nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n}+\left[\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\right]\mathbf{u}=0. $$

This equation is remarkable, because at first sight it seems crazy. The term $\left[\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\right]\mathbf{u}$ involves second derivatives of $\mathbf{u}$ and first derivatives of $\nabla_{\mathbf{n}}:$

$$ \left[\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\right]\mathbf{u}=\nabla_{\mathbf{n}}\nabla_{\mathbf{u}}\mathbf{u}-\nabla_{\mathbf{u}}\nabla_{\mathbf{n}}\mathbf{u}. $$

It thus must depend on how $\mathbf{u}$ and $\mathbf{n}$vary from point to point. But the relative acceleration it produces, $\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n},$ is known to depend only on the values of $\mathbf{u}$ and $\mathbf{n}$ at the fiducial point, not on how $\mathbf{u}$ and $\mathbf{n}$ vary (see box 11.2 F).

I do not understand what this means. The relative acceleration is derived by multiple parallel transport steps, and taking differences of vectors at different locations. That is, its establishment involves more than simply the values of $\mathbf{u}$ and $\mathbf{n}$ at one point. Therefore, if we interpret "the fiducial point" to mean, $\mathscr{Q}$ alone, the statement doesn't make sense to me, at all.

Even if we allow "the fiducial point" to mean any arbitrary point on the fiducial geodesic, the derivation still involves points not on that geodesic, and not determined solely by parallel transport along the fiducial geodesic.

What does it mean to say "$\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n},$ is known to depend only on the values of $\mathbf{u}$ and $\mathbf{n}$ at the fiducial point, not on how $\mathbf{u}$ and $\mathbf{n}$ vary"?

PS: I believe the essential conclusion is that relative acceleration can be expressed as the operation of a multilinear form (tensor field $\mathbf{\text{Riemann}}$) on $\mathbf{u}$ and $\mathbf{n}$. What I'm not getting is the reasoning leading to that conclusion.

The situation seems similar to that of the differential of a multivariable mapping being a linear mapping associated with its derivative matrix.

enter image description here


Just means it's linear in u and n.


Your relative acceleration vector is just geodesic deviation. It is the physical manifestation of the riemann tensor R(u, n)u which is a vactor valued 2-form. It tells you the deviation of a geodesic in the neighborhood of n connected by u, at each point p(t) where t is the affine parameter corresponding to the proper time. Iff the rieamann tensor is nonzero inertial geodesics will accelerate with respect to each other (tidal effect). So you just want the deviation at a given point, not the deviation of the deviation. That's why the tensor contains no derivatives of u and n. So R(lu, n) u =lR(u, n) u where l is a scalar function.

  • $\begingroup$ That's the short answer. I guess I am supposed to understand that the multilinear mapping itself is determined by "how $\mathbf{u}$ and $\mathbf{n}$ vary from point to point", but the value of the mapping does not. If this is the intent of the authors, I must protest that they do not explain themselves well. If their intent was to get me study the discussion carefully, they certainly accomplished that. $\endgroup$ Feb 1 '20 at 23:34
  • $\begingroup$ A vector field can vary from point to point. But it is the same vector field. $\endgroup$
    – Kugutsu-o
    Feb 2 '20 at 10:29
  • $\begingroup$ I love the style of this book. But the worst thing about it is the very goal the authors had, to show of all the different notations and formalism can complement each other in calculations and conceptual understanding. While such an assumption holds some validity it does fail here because you first need to really understand one way really well but the book keeps skipping from one to another casually. $\endgroup$
    – Kugutsu-o
    Feb 2 '20 at 10:35
  • $\begingroup$ If my understanding is correct, the relative acceleration vector is the result of a 3-covariant, 1-contravariant tensor operating twice on $\mathbf{u}$, and once on $\mathbf{n}$. Where $\mathbf{u}$ and $\mathbf{n}$ are the values at $\mathscr{M}$. The tensor itself is dependent on how these (or equivalent) field vectors change from point to point. The individual components of the tensor do depend on how the vectors vary from point to point. That, to me, is no more or less "crazy" than making the same observation about the "tidal forces" term. $\endgroup$ Feb 2 '20 at 18:00
  • $\begingroup$ I told you what it means that it doesn't depend on how the fields vary. A covariant derivative with respect to the vector field n does not depend on how n varies. It contains no derivatives of n. Same for the riemann tensor. It can be proven in 3 lines for each slot using some algebra.don't think you can get it from a picture $\endgroup$
    – Kugutsu-o
    Feb 2 '20 at 18:07

I've already accepted an answer, and don't intend to change that choice. Nonetheless, I have given this some more thought, and have come up with a heuristic way of explaining what the authors may have meant by

But the relative acceleration it produces, $\nabla_{\mathbf{u}}\nabla_{\mathbf{u}}\mathbf{n},$ is known to depend only on the values of $\mathbf{u}$ and $\mathbf{n}$ at the fiducial point, not on how $\mathbf{u}$ and $\mathbf{n}$ vary.

First, using coordinate pairs $\{n,\lambda\},$ establish the unit selector vector


and the tangent vector


This characterization of $\mathbf{u}$ does not address the "scale factor" proposed in Box 11.2.

Consider the vectors



to be "fixed" once and for all during this exercise. By "fixed", I mean that we could cut toothpicks to their lengths and only slide them from point to point by parallel transport. Their values are determined at the coordinate point $\left\{n,\lambda\right\}_\mathscr{M}=\left\{0,0\right\}$ by the coordinate displacements

$$\Delta n=n_\mathscr{Q}-n_\mathscr{M}=n_\mathscr{Q}$$

$$\Delta \lambda=\lambda_\mathscr{R}-\lambda_\mathscr{M}=\lambda_\mathscr{R}$$

The field vectors $\mathbf{n}_f=\mathbf{n}\Delta{n}$ and $\mathbf{u}_f=\frac{d\mathscr{P}}{d\lambda}$ vary from point to point during the exercise. Indeed the entire discussion is about how these vary near $\mathscr{M}$. The standard of measure used to quantify those changes is the pair of vectors $\mathbf{n}_o$ and $\mathbf{u}_o$; which are held as "fixed" as possible by parallel transport.

The situation is similar to the case of the differential of a mapping $\vec{f}:\mathbb{R}^n\to\mathbb{R}^m$ and $\vec{\mathrm{p}}\in\mathbb{R}^n$ in which $$\Delta \vec{f}_\vec{\mathrm{p}}\left[\Delta \vec{\mathrm{p}}\right] =\vec{f}\left[\vec{\mathrm{p}}+\Delta \vec{\mathrm{p}}\right]-\vec{f}\left[\vec{\mathrm{p}}\right] \approx d \vec{f}_\vec{\mathrm{p}}\left[\Delta \vec{\mathrm{p}}\right];$$

where the differential $d \vec{f}_\vec{\mathrm{p}}$ is the linear mapping with the associated derivative (or Jacobian) matrix

$$\left.\frac{d \vec{f}}{d \vec{\mathrm{p}}}\right\rvert_\vec{\mathrm{p}}.$$

The unit selector $\mathbf{n}$ is (aparently) spacelike. Since there is no guarantee two geodesic worldlines will maintain the same relative spacelike distance over time, the physical length of $\mathbf{n}$ might change with changing values of the affine parameter $\lambda$. $\mathbf{u},$ on the otherhand, will always be the time between subsequent ticks of the worldline clock where it is determined.

This is an effort to derive a trilinear expression reflecting the discussion in the book. The underlined expressions are to be held constant under $\nabla$. The postfix dot products contract on the components of $\nabla$ rather than those of the vector being differentiated. $\dagger_{1}$ The equation $\nabla\left[\mathfrak{u}\right]_\mathfrak{u}=0$ is the condition of a geodesic. $\dagger_{2}$ I am assuming coordinate curves of constant $n$ intersect those of constant $\lambda$ so that $\nabla\left[n^{\beta}\right]\cdot\mathfrak{u}=0$. I believe my derivation is correct, but I wouldn't take it to the bank without verifying it. enter image description here

But this only scratches the surface of Box 11.2. To the tune of Ozzy's Mr. Crowley:

Dr. Wheeler
Was it polemically sent
I want to know what you meant
I want to know
I want to know what you meant, yeah

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